# Math Help - [SOLVED] Quotient problem.

1. ## [SOLVED] Quotient problem.

Can someone check where i have gone wrong in this.
$(\frac{u}{v})'= \frac{vu'-uv'}{v^2}$

$\frac{x^3y}{1-xy}$

$\frac{ [(1-xy) \cdot (3x^2y)] - [(x^3y) \cdot (-y)]} {(1-xy)^2}$

$= \frac{(3x^2y-3x^3y^2)-(-x^3y^2)}{(1-xy)^2}$

$=\frac{3x^2y-3x^3y^2+x^3y^2}{(1-xy)^2}$

$=\frac{3x^2y-2x^3y^2}{(1-xy)^2}$

2. What is the answer given by the text book? I did a quick calculation and your answer seems to be correct. Factorised, however, it becomes:

$\frac{-x^2(2xy-3)y}{(xy-1)^2}$

Unless it's implicit?

3. Partial derivative with respect to x, holding y constant? Then you would be fine - but probably this is implicit differentiation where y is an implicit function of x.

Edit:

In which case - and just in case a picture helps...

This is similar to the quotient rule but instead it shows the chain rule...

... wrapped inside the product rule...

Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (which is the inner function of the composite and hence subject to the chain rule).

Try drawing the product rule shape with x^3 and y the two balloons up top, to check that I've differentiated it right (with respect to x). And similarly for the product of -x and y.

__________________________________________

Don't integrate - balloontegrate!

http://www.ballooncalculus.org/forum/top.php

Draw balloons with LaTeX: http://www.ballooncalculus.org/asy/doc.html

4. the question just asked to" find all of the first and second partial derivatives
of .."

$f(x,y)= x^3\ln(1-xy)$

So is this implicit? (<--- don't really know what this means either)

But i need to know $f_{xx}$

Can you tell me which way i need to go about it?]

So that original equaiton above i gave was part of the first derivative of ( $f_x = 3x^2\ln(1-xy) - \frac{x^3y}{1-xy}$. I am now trying to find the second order, but thats the last part of the equation i am differentiating , with respect to x.

Does that make sense?

5. If I understand you correctly you want to find:
$
\frac{\partial f}{\partial x} , \frac{\partial^2f}{\partial x^2} , \frac{\partial f}{\partial y} ,\frac{\partial^2 f}{\partial y^2}, \text{ and } \ \frac{\partial^2 f}{\partial y\partial x}
$

Partial derivatives are pretty easy, just hold everything, but the variable you are diffrentiatin with respect to, as constants.
$
\frac{\partial f}{\partial y} = \frac{\partial }{\partial y} \Bigl(x^3\ln(1-xy)\Bigr) = \frac{-x^4}{1-xy}
$

$
\frac{\partial f}{\partial x} = \frac{\partial }{\partial x}\Bigl(x^3\ln(1-xy)\Bigr) = 3x^2\ln(1-xy) - \frac{yx^3}{1-xy}
$

You need to find $f_{xx}$ right ? just take the partial derivative again.

$
\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial}{\partial x}\left(3x^2\ln(1-xy) - \frac{yx^3}{1-xy}\right)
$

Can you take it from here?

EDIT: Sorry I didnt see your last post. Nevermind, I will post again in few minutes

Assuming that I did not screw up:
$
f_{xx} = 6x\ln(1-xy) + \frac{3x^2-3x^3y-3x^2y+2x^3y^2}{(1-xy)^2}
$

6. can you show me how you got your result , i still keep getting somthing different.

7. ok now i am totally lost , what should i be doing? Using quotient rule i come up with the value i first posted with.

Your value is different. I need to find the second order derivative with respect to x......

8. I just used your results from the first post.